Differential Equations Archives

(x -a)2 + (y –a)2 = a2 …………1 differentiating eq 1 with respect to x, we get, \[\begin{array}{*{35}{l}} 2\left( x-a \right)\text{ }+\text{ }2\left( y-a \right)\text{ }dy/dx~=\text{ }0 \\ \Rightarrow...

Prove that \[{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }{{\mathbf{y}}^{\mathbf{2}}}~=\text{ }\mathbf{c}\text{ }{{\left( {{\mathbf{x}}^{\mathbf{2}}}~+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)}^{\mathbf{2}}}~\] is the general solution of differential equation \[({{\mathbf{x}}^{\mathbf{3}}}-\mathbf{3x}{{\mathbf{y}}^{\mathbf{2}}})\text{ }\mathbf{dx}\text{ }=\text{ }\left( {{\mathbf{y}}^{\mathbf{3}}}-\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{y} \right)\text{ }\mathbf{dy},\]where c is a parameter.

CBSE, Class 12, Differential Equations, Miscellaneous, NCERT

Form the differential equation representing the family of curves given by (x – a)^2 + 2y^2 = a^2, where a is an arbitrary constant.

CBSE, Class 12, Differential Equations, Miscellaneous, NCERT

verify that the given function (implicit or explicit) is a solution of the corresponding differential equation (iii) $y=x\sin 3x:\frac{{{d}^{2}}y}{d{{x}^{2}}}+9y-6\cos 3x=0$ (iv) ${{x}^{2}}=2{{y}^{2}}\log y:({{x}^{2}}+{{y}^{2}})\frac{dy}{dx}-xy=0$

CBSE, Class 12, Differential Equations, Miscellaneous, NCERT

verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.(i)$xy=a{{e}^{x}}+b{{e}^{-x}}+{{x}^{2}}:x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}-xy+{{x}^{2}}-2=0$ (ii) $y={{e}^{x}}(a\cos x+b\sin x):\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}+2y=0$

CBSE, Class 12, Differential Equations, Miscellaneous, NCERT

indicate its order and degree (if defined). $\frac{{{d}^{4}}y}{d{{x}^{4}}}-\sin (\frac{{{d}^{3}}y}{d{{x}^{3}}})=0$

CBSE, Class 12, Differential Equations, Miscellaneous, NCERT

indicate its order and degree (if defined). (i) $\frac{{{d}^{2}}y}{d{{x}^{2}}}+5x(\frac{dy}{dx})-6y=\log x$ (ii) ${{\frac{dy}{dx}}^{3}}-4{{\frac{dy}{dx}}^{2}}+7y=\sin x$

CBSE, Class 12, Differential Equations, Miscellaneous, NCERT

Therefore, its degree is three.

The Integrating Factor of the differential equation $(1-{{y}^{2}})\frac{dx}{dy}+yx=ay(-1

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

Therefore, the correct option is OPTION(D)

The Integrating Factor of the differential equation is A. e–x B. e–y C. 1/x D. x

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

Therefore, the correct option is OPTION(C).

Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

Thus, equation (2) becomes: \[\begin{array}{*{35}{l}} 0\text{ }+\text{ }2\text{ }-\text{ }4\text{ }=\text{ }C\text{ }{{e}^{0}} \\ \Rightarrow ~C\text{ }=\text{ }-2 \\ \end{array}\] Substituting C...

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

Since, the curve passes through origin. Thus, equation 2 becomes 1 = C Substituting C = 1 in equation 2, we get, \[x\text{ }+\text{ }y\text{ }+\text{ }1\text{ }=\text{ }{{e}^{x}}\] Therefore, the...

find a particular solution satisfying the given condition: $\frac{dy}{dx}-3y\cot x=\sin 2x;y=2whenx=\frac{\pi }{2}$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

find a particular solution satisfying the given condition: $(1+{{x}^{2}})\frac{dy}{dx}+2xy=\frac{1}{1+{{x}^{2}}};y=0whenx=1$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

find a particular solution satisfying the given condition: $\frac{dy}{dx}+2y\tan x=\sin x;y=0whenx=\frac{\pi }{3}$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

find the general solution: $(x+3{{y}^{2}})\frac{dy}{dx}=y(y>0)$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

⇒ x = 3y2 + Cy

find the general solution: $ydx+(x-{{y}^{2}})dy=0$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

$x=\frac{{{y}^{3}}}{3}+c$ $x=\frac{{{y}^{2}}}{3}+\frac{c}{y}$

find the general solution: ${{\cos }^{2}}x\frac{dy}{dx}+y=\tan x(0\le x<\frac{\pi }{2})$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

find the general solution: $\frac{dy}{dx}+(\sec x)y=\tan x(0\le x<\frac{\pi }{2})$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

find the general solution: $\frac{dy}{dx}+\frac{y}{x}={{x}^{2}}$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

find the general solution: $\frac{dy}{dx}+3y={{e}^{-2x}}$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

find the general solution: $\frac{dy}{dx}+2y=\sin x$

CBSE, Class 12, Differential Equations, Exercise 9.6, NCERT

Which of the following is a homogeneous differential equation? A. (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 B. (x y) dx – (x^3 + y^3) dy = 0 C. (x^3 + 2y^2) dx + 2xy dy = 0 D. y^2dx + (x^2 – x y – y^2) dy = 0

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

D. y2dx + (x2 – x y – y2) dy = 0

A homogeneous differential equation of the from $\frac{dx}{dy}=h(\frac{x}{y})$ can be solved by making the substitution. (A) y = v x (B) v = y x (C) x = v y (D) x = v

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

(C) x = v y

find the particular solution satisfying the given condition: $2xy+{{y}^{2}}-2{{x}^{2}}\frac{dy}{dx}=0;y=2whenx=1$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

The required solution of the differential equation.

find the particular solution satisfying the given condition: $\frac{dy}{dx}-\frac{y}{x}+\cos ec(\frac{y}{x})=0;y=0whenx=1$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

find the particular solution satisfying the given condition: $[x{{\sin }^{2}}x(\frac{y}{x})-y]dx+xdy=0;y=\frac{\pi }{4}whenx=1$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

find the particular solution satisfying the given condition: \[{{\mathbf{x}}^{\mathbf{2}}}\mathbf{dy}\text{ }+\text{ }\left( \mathbf{x}\text{ }\mathbf{y}\text{ }+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)\mathbf{dx}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{1}\text{ }\mathbf{when}\text{ }\mathbf{x}\text{ }=\text{ }\mathbf{1}\]

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

find the particular solution satisfying the given condition: (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them $(1+{{e}^{\frac{x}{y}}})dx+{{e}^{\frac{x}{y}}}(1-\frac{x}{y})dy=0$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them. $ydx+x\log \frac{y}{x}dy-2xdy=0$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them.$x\frac{dy}{dx}-y+x\sin \frac{y}{x}=0$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them.$\{x\cos (\frac{y}{x})+y\sin (\frac{y}{x})\}ydx=\{y\sin (\frac{y}{x})-x\cos (\frac{y}{x})\}xdy$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them. $xdy-ydx=\sqrt{({{x}^{2}}+{{y}^{2}})}dy$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them. ${{x}^{2}}\frac{dy}{dx}={{x}^{2}}-2{{y}^{2}}+xy$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them. \[\left( {{\mathbf{x}}^{\mathbf{2}}}~-\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)\mathbf{dx}\text{ }+\text{ }\mathbf{2xy}\text{ }\mathbf{dy}\text{ }=\text{ }\mathbf{0}\]

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

x2 + y2 = Cx

show that the given differential equation is homogeneous and solve each of them. (x – y) dy – (x + y) dx = 0

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them. $y’=\frac{x+y}{x}$

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

show that the given differential equation is homogeneous and solve each of them. \[~\left( {{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{x}\text{ }\mathbf{y} \right)\text{ }\mathbf{dy}\text{ }=\text{ }\left( {{\mathbf{x}}^{\mathbf{2}}}~+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)\text{ }\mathbf{dx}\]

CBSE, Class 12, Differential Equations, Exercise 9.5, NCERT

The general equation of the differential equation $\frac{dy}{dx}={{e}^{x+y}}is(A){{e}^{x}}+{{e}^{-y}}=c(B){{e}^{x}}+{{e}^{y}}=c(C){{e}^{-x}}+{{e}^{y}}=c(D){{e}^{-x}}+{{e}^{-y}}=c$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

The correct option is option(A) ex + e-y = C

In a culture, the bacteria count is 1, 00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2, 00,000, if the rate of growth of bacteria is proportional to the number present?

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge 2 = 0.6931).

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

For the differential equation $xy\frac{dy}{dx}=(x+2)(y+2)$ ,Find the solution curve passing through the point (1, –1).

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

Find the equation of a curve passing through the point (0, 0) and whose differential equation is $y’={{e}^{x}}\sin x$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find a particular solution Satisfying the given condition: $\frac{dy}{dx}=y\tan x;y=1whenx=0$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find a particular solution Satisfying the given condition: $\cos (\frac{dy}{dx})=a(a\in R);y=1whenx=0$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find a particular solution Satisfying the given condition: $x({{x}^{2}}-1)\frac{dy}{dx}=1;y=0whenx=2$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find a particular solution Satisfying the given condition: $({{x}^{3}}+{{x}^{2}}+x+1)\frac{dy}{dx}=2{{x}^{2}}+x;y=1whenx=0$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: ${{e}^{x}}\tan ydx+(1-{{e}^{x}}){{\sec }^{2}}ydy=0$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: $\frac{dy}{dx}={{\sin }^{-1}}x$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: ${{x}^{5}}\frac{dy}{dx}=-{{y}^{5}}$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: ylogydx-xdy=0

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: $\frac{dy}{dx}=(1+{{x}^{2}})(1+{{y}^{2}})$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: $({{e}^{x}}+{{e}^{-x}})dy-({{e}^{x}}-{{e}^{-x}})dx=0$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: ${{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy=0$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: $\frac{dy}{dx}+y=1(y\ne 1)$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution: $\frac{dy}{dx}=\sqrt{(4-{{y}^{2}})}(-2

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

find the general solution:$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

CBSE, Class 12, Differential Equations, Exercise 9.4, NCERT

Which of the following differential equations has y = x as one of its particular solution? $(A)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}-{{x}^{2}}\frac{dy}{dx}+xy=x(B)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}+{{x}^{{}}}\frac{dy}{dx}+xy=x(C)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}-{{x}^{2}}\frac{dy}{dx}+xy=0(D)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}+{{x}^{{}}}\frac{dy}{dx}+xy=0$

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

= 0 – (x2 × 1) + (x × x) = -x2 + x2 = 0

Which of the following differential equations has y = c1 ex + c2 e-x as the general solution? $(A)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}+y=0(B)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}-y=0(C)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}+1=0(D)\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}-1=0$.

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

[From eq. (i)] Therefore,The correct option is option (B) .

Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

Centre of the circle on axis is . Equation of the circle having centre on axis an radius unit is ……….(i) Here is the only arbitrary constant, therefore we will differentiate only once. ...

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

\[\begin{array}{*{35}{l}} \Rightarrow ~x\text{ }{{\left( y ‘\right)}^{2}}~+\text{ }xyy’’\text{ }-\text{ }yy’\text{ }=\text{ }0 \\ \Rightarrow ~xyy’’\text{ }+\text{ }x{{\left( y’...

Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

\[\begin{array}{*{35}{l}} \Rightarrow ~-x\text{ }{{\left( y’ \right)}^{2}}~-\text{ }xyy”\text{ }+\text{ }yy’\text{ }=\text{ }0 \\ \Rightarrow ~xyy”\text{ }+\text{ }x\text{ }{{\left( y’...

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

\[{{x}^{2}}~=\text{ }4ay\text{ }\ldots \text{ }[equation\text{ }\left( i \right)\]

Form the differential equation of the family of circles touching the y-axis at origin.

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

Solution: assuming (p, 0) be the centre of the circle. Therefore, it touches the y – axis at origin, its radius is p. Since, the equation of the circle with centre (p, 0) and radius (p) is...

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b in \[\mathbf{y}\text{ }=\text{ }{{\mathbf{e}}^{\mathbf{x}}}~\left( \mathbf{a}\text{ }\mathbf{cos}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{b}\text{ }\mathbf{sin}\text{ }\mathbf{x} \right)\]

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

Differentiating both sides two times w.r.t. [By eq. (i)]……….(ii) Again differentiating w.r.t. , [By eq. (i)]

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b in \[~\mathbf{y}\text{ }=\text{ }{{\mathbf{e}}^{\mathbf{2x}}}~\left( \mathbf{a}\text{ }+\text{ }\mathbf{bx} \right)\]

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

Differentiating both sides two times w.r.t. [By eq. (i)]……….(ii) Again differentiating w.r.t. , ……….(iii) Now from eq. (ii), Putting this value of in eq. (iii), ...

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b in \[\mathbf{y}\text{ }=\text{ }\mathbf{a}{{\mathbf{e}}^{\mathbf{3x}}}~+\text{ }\mathbf{b}{{\mathbf{e}}^{-\mathbf{2x}}}\]

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

DifferentiatING both sides two times w.r.t. ……….(ii) Again differentiating w.r.t. , ……….(iii) Multiplying eq. (i) by 3 and subtracting eq. (ii) from it, we get ……….(iv) Again multiplying eq....

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b in \[{{\mathbf{y}}^{\mathbf{2}}}~=\text{ }\mathbf{a}\text{ }({{\mathbf{b}}^{\mathbf{2}}}~-\text{ }{{\mathbf{x}}^{\mathbf{2}}})\]

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

Equation of the family of curves ……….(i) Differentiating both sides two times w.r.t. ……….(ii) Again differentiating w.r.t. , ……….(iii) Putting this value of in eq. (ii), we get ...

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b in $\frac{x}{a}+\frac{y}{b}=1$

CBSE, Class 12, Differential Equations, Exercise 9.3, NCERT

Equation of the family of curves ……….(i) Differentiating both sides two times w.r.t. ……….(ii) Again differentiating w.r.t. , Multiplying both sides by , which the required differential...

The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

The number of arbitrary constants in a particular solution of a differential equation of any order is zero (0) as a particular solution is a solution which contains no arbitrary constant....

The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) 2 (C) 3 (D) 4

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

The correct option is Option (D) . Since, the number of arbitrary constants ( etc.) in the general solution of a differential equation of order is

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: $y=\sqrt{({{a}^{2}}-{{x}^{2}})},\in (-a,a):x+y\frac{dy}{dx}=0(y\ne 0)$

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

From eq. (i), = = ……….(iii) Putting the values of and from eq. (i) and (iii) in L.H.S. of eq. (ii), = = = R.H.S. of eq. (ii) Hence, Function given by eq. (i) is a solution...

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \[\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\mathbf{y}\text{ }:\text{ }{{\mathbf{y}}^{\mathbf{2}}}~\mathbf{y}\prime \text{ }+\text{ }{{\mathbf{y}}^{\mathbf{2}}}~+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

Differentiating both sides of eq. (i) w.r.t we have = eq. (ii) Hence, Function given by eq. (i) is a solution of

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y – cos y = x : (y sin y + cos y + x) y′ = y

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

Differentiating both sides of eq. (i) w.r.t we have ……….(iii) Putting the value of from eq. (i) and value of from eq. (iii) in L.H.S. of eq. (ii), = R.H.S. of (ii) Hence, Function given by...

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: $xy=\log y+c:y’=\frac{{{y}^{2}}}{1-xy}(xy\ne 1)$

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

Differentiating both sides of eq. (i) w.r.t we have Hence, Function (implicit) given by eq. (i) is a solution of .

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \[\mathbf{y}\text{ }=\text{ }\mathbf{x}\text{ }\mathbf{sinx}\text{ }:\text{ }\mathbf{xy}\text{ }=\text{ }\mathbf{y}\text{ }+\text{ }\mathbf{x}\text{ }(\surd ({{\mathbf{x}}^{\mathbf{2}}}~-\text{ }{{\mathbf{y}}^{\mathbf{2}}}))\text{ }\left( \mathbf{x}\text{ }\ne \text{ }\mathbf{0}\text{ }\mathbf{and}\text{ }\mathbf{x}>\mathbf{y}\text{ }\mathbf{or}\text{ }\mathbf{x}<\text{ }\text{ }-\mathbf{y} \right)\]

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

..(ii) = L.H.S. of eq. (ii) = R.H.S. of eq. (ii) = = [From eq. (i)] = = = = = L.H.S. = R.H.S Hence, given by eq. (i) is a solution of .

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:y = Ax : xy′ = y (x ≠ 0)

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

Differentiating both sides with respect to x, we get, y’ = A Substituting the values of y’ in the given differential equations, we get, \[\begin{array}{*{35}{l}} =\text{ }xy \\ =\text{ }x\text{...

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:\[\mathbf{y}\text{ }=\text{ }\surd \left( \mathbf{1}\text{ }+\text{ }{{\mathbf{x}}^{\mathbf{2}}} \right):\text{ }\mathbf{y}\text{ }=\text{ }(\left( \mathbf{xy} \right)/(\mathbf{1}\text{ }+\text{ }{{\mathbf{x}}^{\mathbf{2}}}))\]

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y = cos x + C : y′ + sin x = 0

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

Differentiating both sides with respect to x, we get, y’ = -sinx Then, Substituting the values of y’ in the given differential equations, we get, \[\begin{array}{*{35}{l}} =\text{ }y\text{ }+\text{...

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \[~\mathbf{y}\text{ }=\text{ }{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{C}\text{ }:\text{ }\mathbf{y}\prime \text{ }\text{ }-\mathbf{2x}\text{ }\text{ }-\mathbf{2}\text{ }=\text{ }\mathbf{0}\]

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

Differentiating both sides with respect to x, we get, y’ = 2x + 2 Substituting the values of y’ in the given differential equations, we get, \[\begin{array}{*{35}{l}} =\text{ }y\text{ }-\text{...

verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y = e^x + 1 : y″ – y′ = 0

CBSE, Class 12, Differential Equations, Exercise 9.2, NCERT

Differentiating both sides with respect to x, we get, ⇒ y” = ex Then, Substituting the values of y’ and y” in the given differential equations, we get, y” – y’ = ex – ex = RHS. Therefore, the given...

Determine order and degree (if defined) of differential equations given in The order of the differential equation $2{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-3\frac{dy}{dx}+y=0$is: (A) 2 (B) 1 (C) 0 (D) Not defined

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 2. Therefore, The correct option is option (A) .

Determine order and degree (if defined) of differential equations given in ${{(\frac{{{d}^{2}}y}{d{{x}^{2}}})}^{3}}+{{(\frac{dy}{dx})}^{2}}+\sin (\frac{dy}{dx})+1=0$ is: (A) 3 (B) 2 (C) 1 (D) Not defined

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

Given: ……….(i) This equation is not a polynomial in derivatives as is a T-function of derivative Therefore, degree of given equation is not defined. Hence, The correct option is option...

Determine order and degree (if defined) of differential equations given in \[\mathbf{y’’’}\text{ }+\text{ }\mathbf{2y’}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{0}\]

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 2. The given differential equation is a polynomial equation in derivative and . It may be noted that is not a...

Determine order and degree (if defined) of differential equations given in \[\mathbf{y’’’}\text{ }+\text{ }{{\left( \mathbf{y’} \right)}^{\mathbf{2}}}~+\text{ }\mathbf{2y}\text{ }=\text{ }\mathbf{0}\]

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 2. The given differential equation is a polynomial equation in derivatives and and the highest power raised...

Determine order and degree (if defined) of differential equations given in $y’+y={{e}^{x}}$

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 1. The given differential equation is a polynomial equation in derivative . It may be noted that is an...

Determine order and degree (if defined) of differential equations given in y’’’ + 2y’’ + y’ = 0

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 3. The given differential equation is a polynomial equation in derivatives and and the highest power raised...

Determine order and degree (if defined) of differential equations given in \[{{\left( \mathbf{y”’} \right)}^{\mathbf{2}}}~+\text{ }{{\left( \mathbf{y”} \right)}^{\mathbf{3}}}~+\text{ }{{\left( \mathbf{y’} \right)}^{\mathbf{4}}}~+\text{ }{{\mathbf{y}}^{\mathbf{5}}}~=\text{ }\mathbf{0}\]

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 3. The given differential equation is a polynomial equation in derivatives and the highest power raised to...

Determine order and degree (if defined) of differential equations given in $\frac{{{d}^{2}}y}{d{{x}^{2}}}=\cos 3x+\sin 3x$

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 2. The given differential equation is a polynomial equation in derivatives and the highest power raised to...

Determine order and degree (if defined) of differential equations given in ${{(\frac{{{d}^{2}}y}{d{{x}^{2}}})}^{2}}+\cos (\frac{dy}{dx})=0$

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 2. The given differential equation is not a polynomial equation in derivatives as the term is a T-function of...

Determine order and degree (if defined) of differential equations given in $({{\frac{ds}{dt}}^{4}})+3s\frac{{{\partial }^{2}}s}{\partial {{t}^{2}}}=0$

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The highest order derivative present in the differential equation is and its order is 1. The given differential equation is a polynomial equation in derivative and the highest power raised to...

Determine order and degree (if defined) of differential equations given in y’+5y=0

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

Since, the highest order derivative present in the differential equation is y’, so its order is one. Therefore, the given differential equation is a polynomial equation in its derivatives. So, its...

Determine order and degree (if defined) of differential equations given in $\frac{{{d}^{4}}y}{d{{x}^{^{4}}}}+\sin (y”)=0$

CBSE, Class 12, Differential Equations, Exercise 9.1, NCERT

The given differential equation is, ⇒ y”” + sin (y’’’) = 0 The highest order derivative present in the differential equation is y’’’’,Therefore, its order is three. Hence, the given differential...