Differential Equations — Page 13 | JEE Mathematics

Q121

Let

y=y(x)

be the solution of the differential equation

(x^2-3y^2)dx+3xy~dy=0,y(1)=1

. Then

6y^2(e)

is equal to

A

\dfrac{3}{2}\mathrm{e}^2

B

3\mathrm{e}^2

C

\mathrm{e}^2

D

2\mathrm{e}^2

Correct Answer

Option D

Solution

Given,

\left( {{x^2} - 3{y^2}} \right)dx + 3xydy = 0

\Rightarrow {x^2}dx - 3{y^2}dx + 3xydy = 0

\Rightarrow {{{x^2}dx} \over {3xdx}} - {{3{y^2}dx} \over {3xdx}} + {{3ydy} \over {3xdx}} = 0

\Rightarrow {x \over 3} - {{{y^2}} \over x} + y{{dy} \over {dx}} = 0

Let

{y^2} = t

2y{{dy} \over {dx}} = {{dt} \over {dx}}

$\therefore$

{1 \over 2}{{dt} \over {dx}} - {t \over x} + {x \over 3} = 0

\Rightarrow {{dt} \over {dx}} - {{2t} \over x} = - {{2x} \over 3}

This is linear Differential Equation.

\text{ I.F. }=e^{\int-\frac{2}{x} d x}=e^{-2 \ln |x|}=e^{\ln \frac{1}{x^2}}=\frac{1}{x^2}

So, $t \cdot \dfrac{1}{x^2}=\int \dfrac{1}{x^2}\left(\dfrac{-2 x}{3}\right) d x$ $\Rightarrow \dfrac{y^2}{x^2}=\dfrac{-2}{3} \ln |x|+C$ When, $x=1, y=1$

1=\frac{-2}{3} \ln (1)+C \Rightarrow C=1

\therefore \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+1

At $x=e, \dfrac{y^2(e)}{e^2}=\dfrac{-2}{3}+1$ $\Rightarrow y^2(e)=\dfrac{e^2}{3}$ $\Rightarrow 6 y^2(e)=2 e^2$

Q122

The solution curve, of the differential equation

2 y \dfrac{\mathrm{d} y}{\mathrm{~d} x}+3=5 \dfrac{\mathrm{d} y}{\mathrm{~d} x}

, passing through the point

(0,1)

is a conic, whose vertex lies on the line :

A

2 x+3 y=-9

B

2 x+3 y=-6

C

2 x+3 y=9

D

2 x+3 y=6

Correct Answer

Option C

Solution

\begin{aligned} & 2 y \frac{d y}{d x}+3=5 \frac{d y}{d x} \\ & 2 y d y+3 d x=5 d y \\ & y^2+3 x=5 y+\left.c\right|_{(0,1)} \\ & 1+0=5+c \\ & c=-4 \\ & y^2-5 y=-3 x-4 \\ & y^2-5 y+\frac{25}{4}-\frac{25}{4}=-3 x-4 \\ & \left(y-\frac{5}{2}\right)^2=-3 x+\frac{9}{4} \\ & \left(y-\frac{5}{2}\right)^2=-3\left(x-\frac{3}{4}\right) \\ & \left(\frac{3}{4}, \frac{5}{2}\right) \text{ satisfies by } 2 x+3 y=9 \end{aligned}

Q123

Let

x=x(y)

be the solution of the differential equation

2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0, y>-1

with

x\left(e^{4}-2\right)=1

. Then

x\left(e^{9}-2\right)

is equal to :

A

\dfrac{4}{9}

B

\dfrac{32}{9}

C

\dfrac{10}{3}

D 3

Correct Answer

Option B

Solution

\begin{aligned} & 2(y+2) \ln (y+2) d x+(x+4-2 \ln (y+2)) d y=0 \\\\ & 2 \ln (y+2)+(x+4-2 \ln (y+2)) \frac{1}{y+2} \cdot \frac{d y}{d x}=0 \\\\ & \text{ let, } \ln (y+2)=t \\\\ & \frac{1}{y+2} \cdot \frac{d y}{d x}=\frac{d t}{d x} \\\\ & 2 t+(x+4-2 t) \cdot \frac{d t}{d x}=0 \\\\ & (x+4-2 t) \frac{d t}{d x}=-2 t \\\\ & \frac{d x}{d t}=\frac{2 t-4-x}{2 t} \\\\ & \frac{d x}{d t}+\frac{x}{2 t}=\frac{2 t-4}{2 t} \end{aligned}

\begin{aligned} & x \cdot t^{1 / 2}=\int \frac{2 t-4}{2 t} \cdot t^{1 / 2} \cdot d t \\\\ & x \cdot t^{1 / 2}=\int\left(t^{1 / 2}-\frac{2}{t^{1 / 2}}\right) \cdot d t \\\\ & =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-2 \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\\\ & x \cdot t^{\frac{1}{2}}=\frac{2 t^{\frac{3}{2}}}{3}-4 \mathrm{t}^{\frac{1}{2}}+C \\\\ & x=\frac{2}{3} \cdot \mathrm{t}-4+\mathrm{C} \cdot \mathrm{t}^{\frac{-1}{2}} \\\\ & x=\frac{2}{3} \ln (\mathrm{y}+2)-4+C \cdot(\ln (\mathrm{y}+2))^{\frac{-1}{2}} \end{aligned}

Put $y=e^4-2, x=1$

\begin{aligned} & 1=\frac{2}{3} \times 4-4+C \times \frac{1}{2} \\\\ & \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \\\\ & C=\frac{14}{3} \\\\ & x=\frac{2}{3} \times 9-4+\frac{14}{3} \times \frac{1}{3} \\\\ & =2+\frac{14}{9} \\\\ & =\frac{32}{9} \end{aligned}

Q124

Let

y=y_{1}(x)

and

y=y_{2}(x)

be the solution curves of the differential equation

\dfrac{d y}{d x}=y+7

with initial conditions

y_{1}(0)=0

and

y_{2}(0)=1

respectively. Then the curves

y=y_{1}(x)

and

y=y_{2}(x)

intersect at

A no point

B two points

C infinite number of points

D one point

Correct Answer

Option A

Solution

The given differential equation is

\frac{d y}{d x} = y + 7

This is a first order linear differential equation and can be solved using an integrating factor.

Rearrange the equation to the standard form of a linear differential equation :

\frac{d y}{d x} - y = 7

The integrating factor is $e^{-\int dx} = e^{-x}$ .

Multiplying each side of the equation by the integrating factor gives :

e^{-x} \frac{d y}{d x} - e^{-x} y = 7e^{-x}

The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$ .

So we can write the equation as :

\frac{d}{d x}(e^{-x}y) = 7e^{-x}

Integrate both sides with respect to $x$ :

e^{-x}y = -7e^{-x} + C

Multiply both sides by $e^{x}$ to isolate $y$ :

y = -7 + Ce^{x}

So, the general solution to the differential equation is $y = -7 + Ce^{x}$ .

Now, let's apply the initial conditions to find the particular solutions : For $y_{1}(0)=0$ , we substitute into the general solution and solve for $C$ :

0 = -7 + C

So, $C = 7$ , and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$ .

For $y_{2}(0)=1$ , again substitute into the general solution:

1 = -7 + C

So, $C = 8$ , and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$ .

The two curves intersect when $y_{1}(x) = y_{2}(x)$ .

Setting these equal and solving for $x$ gives :

7e^{x} - 7 = 8e^{x} - 7

e^{x} = 0

But $e^{x} = 0$ has no solution, because the exponential function never equals zero.

So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.

Q125

Let

y=y(x), y > 0

, be a solution curve of the differential equation

\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x

. If

y(0)=1

and

y(2 \sqrt{2})=\beta

, then

A

e^{\beta^{-1}}=e^{-2}(3+2 \sqrt{2})

B

e^{3 \beta^{-1}}=e(5+\sqrt{2})

C

e^{3 \beta^{-1}}=e(3+2 \sqrt{2})

D

e^{\beta^{-1}}=e^{-2}(5+\sqrt{2})

Correct Answer

Option C

Solution

\begin{aligned} & \left(1+x^2\right) d y=y(x-y) d x \\\\ & y(0)=1 \cdot y(2 \sqrt{2})=\beta \\\\ & \frac{d y}{d x}=\frac{y x-y^2}{1+x^2} \\\\ & \frac{d y}{d x}+y\left(\frac{-x}{1+x^2}\right)=\left(\frac{-1}{1+x^2}\right) y^2 \\\\ & \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{y}\left(\frac{-x}{1+x^2}\right)=\frac{-1}{1+x^2} \\\\ & \text{ put } \frac{1}{y}=t \text{ then } \frac{-1}{y^2} \frac{d y}{d x}=\frac{d t}{d x} \end{aligned}

\frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \frac{\mathrm{x}}{1+\mathrm{x}^2}=\frac{1}{1+\mathrm{x}^2}

\text{ I.F }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\frac{1}{2} \ln \left(1+\mathrm{x}^2\right)}=\sqrt{1+\mathrm{x}^2}

t \sqrt{1+x^2}=\int \frac{\sqrt{1+x^2}}{1+x^2} d x

\begin{aligned} & \frac{1}{y} \sqrt{1+x^2}=\int \frac{1}{\sqrt{1+x^2}} d x \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+C \\\\ & \because y(0)=1 \Rightarrow C=1 \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+1 \\\\ & \text{ For } y=2 \sqrt{2} \\\\ & \frac{3}{y}=\ln |2 \sqrt{2}+3|+1 \end{aligned}

\begin{gathered} y=\beta=\frac{3}{1+\ln |2 \sqrt{2}+3|} \\\\ \Rightarrow 3 \beta^{-1}=1+\ln |2 \sqrt{2}+3| \end{gathered}

Isolating the term $e^{3 \beta^{-1}}$ , we get :

e^{3 \beta^{-1}} = e^{1+\ln |2 \sqrt{2}+3|}.

This can be simplified using the rule $e^{a+b} = e^a e^b$ to :

e^{3 \beta^{-1}} = e \cdot e^{\ln |2 \sqrt{2}+3|}.

Since $e^{\ln(x)} = x$ for any $x$ , this simplifies to :

e^{3 \beta^{-1}} = e |2 \sqrt{2}+3|.

Using the given value of $\beta$ , which is $\dfrac{3}{1+\ln |2 \sqrt{2}+3|}$ , we find :

e^{3 \beta^{-1}} = e |2 \sqrt{2}+3| = e (2\sqrt{2} + 3),

Since $2\sqrt{2}+3$ is positive and so the absolute value does not affect the result.

Q126

Let

y=y(x)

be the solution of the differential equation

\dfrac{d y}{d x}+\dfrac{5}{x\left(x^{5}+1\right)} y=\dfrac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0

. If

y(1)=2

, then

y(2)

is equal to :

A

\dfrac{693}{128}

B

\dfrac{697}{128}

C

\dfrac{637}{128}

D

\dfrac{679}{128}

Correct Answer

Option A

Solution

I.F $=\mathrm{e}^{\int \dfrac{5 \mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^5+1\right)}}=\mathrm{e}^{\int \dfrac{5 \mathrm{x}^{-6} \mathrm{dx}}{\left(\mathrm{x}^{-5}+1\right)}}$ Put, $1+\mathrm{x}^{-5}=\mathrm{t} \Rightarrow-5 \mathrm{x}^{-6} \mathrm{dx}=\mathrm{dt}$ $\therefore$

e^{\int-\frac{d t}{t}}=e^{-\ln t}=\frac{1}{t}=\frac{x^5}{1+x^5}

\begin{aligned} y \cdot \frac{x^5}{1+x^5} & =\int \frac{x^5}{\left(1+x^5\right)} \times \frac{\left(1+x^5\right)^2}{x^7} d x \\\\ & =\int x^3 d x+\int x^{-2} d x \end{aligned}

y \cdot \frac{x^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+c

\text{ Given that: } x=1 \Rightarrow y=2

\begin{aligned} & 2 \cdot \frac{1}{2}=\frac{1}{4}-1+\mathrm{c} \\\\ & \mathrm{c}=\frac{7}{4} \\\\ & \mathrm{y} \cdot \frac{\mathrm{x}^5}{1+\mathrm{x}^5}=\frac{\mathrm{x}^4}{4}-\frac{1}{\mathrm{x}}+\frac{7}{4} \\\\ & \text{ Now put, } \mathrm{x}=2 \\\\ & \mathrm{y} \cdot\left(\frac{32}{33}\right)=\frac{21}{4} \\\\ & \mathrm{y}=\frac{693}{128} \end{aligned}

Q127

Let

y=y(x)

be a solution curve of the differential equation.

\left(1-x^{2} y^{2}\right) d x=y d x+x d y

. If the line

x=1

intersects the curve

y=y(x)

at

y=2

and the line

x=2

intersects the curve

y=y(x)

at

y=\alpha

, then a value of

\alpha

is :

A

\dfrac{1+3 e^{2}}{2\left(3 e^{2}-1\right)}

B

\dfrac{3 e^{2}}{2\left(3 e^{2}-1\right)}

C

\dfrac{1-3 e^{2}}{2\left(3 e^{2}+1\right)}

D

\dfrac{3 e^{2}}{2\left(3 e^{2}+1\right)}

Correct Answer

Option A

Solution

We have,

\begin{aligned} & \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\\\ & d x=\frac{y d x+x d y}{1-(x y)^2} \end{aligned}

On integrating both sides, we get

\begin{aligned} \int d x & =\int \frac{d(x y)}{1-(x y)^2} \\\\ x & =\frac{1}{2} \log \left|\frac{1+x y}{1-x y}\right|+C \end{aligned}

As, $y(1)=2$

\begin{aligned} 1 & =\frac{1}{2} \log \left|\frac{1+2}{1-2}\right|+C \\\\ \Rightarrow C & =1-\frac{1}{2} \log 3 \end{aligned}

Now, substitute $x=2$ as $y(2)=\alpha$

2=\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \log 3

\begin{aligned} 1+\frac{1}{2} \log 3 & =\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\\\ 2+\log 3 & =\log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \end{aligned}

\begin{aligned} &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha} \mid =3 e^2 \\\\ &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha}= \pm \frac{3 e^2}{1} \\\\ & \Rightarrow\frac{1}{2 \alpha} =\frac{ \pm 3 e^2+1}{ \pm 3 e^2-1} \\\\ &\Rightarrow 2 \alpha=\frac{ \pm 3 e^2-1}{ \pm 3 e^2+1} \end{aligned}

\begin{array}{ll} \Rightarrow \alpha=\frac{1}{2}\left(\frac{ \pm 3 e^2-1}{1 \pm 3 e^2}\right) \\\\ \therefore \alpha=\frac{3 e^2-1}{2\left(1+3 e^2\right)} \text{ or } \frac{3 e^2+1}{2\left(3 e^2-1\right)} \end{array}

Q128

Let

f

be a differentiable function such that

{x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt}

,

f(1) = {2 \over 3}

. Then

18f(3)

is equal to :

A 160

B 210

C 150

D 180

Correct Answer

Option A

Solution

Given that

x^2 f(x)-x=4 \int_0^x t f(t) d t

On differentiating both sides with respect to $x$ , we get

\begin{array}{rlrl} & 2 x f(x)+x^2 f^{\prime}(x)-1 =4 x f(x) \\\\ &\Rightarrow x^2 f^{\prime}(x)-2 x f(x)-1 =0 \\\\ &\Rightarrow x^2 \frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\ &\frac{d y}{d x}-\frac{2}{x} y =\frac{1}{x^2} \end{array}

On comparing above equation with $\dfrac{d y}{d x}+P y=Q$ , where $P=-\dfrac{2}{x}, Q=\dfrac{1}{x^2}$ Now, IF $=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2$ Solution is $y\left(\dfrac{1}{x^2}\right)=\int\left(\dfrac{1}{x^2}\right) \dfrac{1}{x^2} d x$

\begin{array}{ll} &\Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4} d x \Rightarrow \frac{y}{x^2}=-\frac{1}{3 x^3}+C \\\\ &\Rightarrow y =-\frac{1}{3 x}+C x^2 \end{array}

Given, $f(1)=\dfrac{2}{3}$ . So, $\dfrac{2}{3}=-\dfrac{1}{3}+C \Rightarrow C=1$ Thus, $y=f(x)=-\dfrac{1}{3 x}+x^2$

\therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160

Q129

If the solution curve

f(x, y)=0

of the differential equation

\left(1+\log _{e} x\right) \dfrac{d x}{d y}-x \log _{e} x=e^{y}, x > 0

, passes through the points

(1,0)

and

(\alpha, 2)

, then

\alpha^{\alpha}

is equal to :

A

e^{\sqrt{2} e^{2}}

B

e^{2 e^{\sqrt{2}}}

C

e^{e^{2}}

D

e^{2 e^{2}}

Correct Answer

Option D

Solution

We have, $\left(1+\log _e x\right) \dfrac{d x}{d y}-x \log _e x=e^{y}, x>0$ Put $x \log _e x=t$

\begin{aligned} & \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \therefore \frac{d t}{d y}-t =e^{y} \end{aligned}

Now, IF $=e^{\int-d y}=e^{-y}$ $\begin{aligned} & \therefore \text{ General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$ Equation (i), passes through the point $(1,0)$

\begin{aligned} & \therefore 0=0+c \\\\ & \Rightarrow c=0 \\\\ & \therefore \left(x \log _e x\right) e^{-y}=y \\\\ & \Rightarrow x \log _e x=y e^y \end{aligned}

Which passes through $(\alpha, 2)$

\begin{aligned} & \therefore \alpha \log _e \alpha=2 e^2 \\\\ & \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\ & \Rightarrow \alpha^\alpha=e^{2 e^2} \end{aligned}

Q130

Let

\alpha

be a non-zero real number. Suppose

f: \mathbf{R} \rightarrow \mathbf{R}

is a differentiable function such that

f(0)=2

and

\lim\limits_{x \rightarrow-\infty} f(x)=1

. If

f^{\prime}(x)=\alpha f(x)+3

, for all

x \in \mathbf{R}

, then

f\left(-\log _{\mathrm{e}} 2\right)

is equal to :

A 7

B 9

C 3

D 5

Correct Answer

Option B

Solution

\begin{aligned} & f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1 \\\\ & f^{\prime}(x)-x \cdot f(x)=3 \\\\ & \text{ I.F }=e^{-\alpha x} \\\\ & y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x \\\\ & f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c \\\\ & x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \\\\ & f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\ & x \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0) \\\\ & \alpha=-3 \therefore c=1 \\\\ & f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\ & =1+e^{3 \ln 2}=9 \end{aligned}