Question

The solution of the differential equation ${\text{ydx - (x + 2}}{{\text{y}}^2}{\text{)dy = 0 }}$is x = f (y). If f (-1) = 1, then f (1) is equal to:
A. 2
B. 3
C. 4
D. 1

Answer 1

Hint: To solve this question, we will find the solution of the given differential equation and put the value in the equation formed to find the integration constant. After it, we will find the required value.

Complete step-by-step answer: -
Now, we are given the equation
${\text{ydx - (x + 2}}{{\text{y}}^2}{\text{)dy = 0 }}$. Now, to simplify this equation, we can write it as,
${\text{ydx = (x + 2}}{{\text{y}}^2}{\text{)dy}}$. Now, dividing both sides by ydy, we get
$\dfrac{{{\text{dx}}}}{{{\text{dy}}}}{\text{ = }}\dfrac{{\text{x}}}{{\text{y}}}{\text{ + 2y}}$
So, the equation is $\dfrac{{{\text{dx}}}}{{{\text{dy}}}}{\text{ - }}\dfrac{{\text{x}}}{{\text{y}}}{\text{ = 2y}}$.
Now, the above equation represents the linear equation of the form $\dfrac{{{\text{dx}}}}{{{\text{dy}}}}{\text{ + Px = Q}}$whose solution can be found by ${\text{x(I}}{\text{.F) = }}\int {{\text{Q(I}}{\text{.F)dy}}}$, where I.F is the integrating factor and ${\text{I}}{\text{.F = }}{{\text{e}}^{\int {{\text{Pdy}}} }}$.
Now, comparing the equation $\dfrac{{{\text{dx}}}}{{{\text{dy}}}}{\text{ - }}\dfrac{{\text{x}}}{{\text{y}}}{\text{ = 2y}}$ with linear form, we get
${\text{P = - }}\dfrac{1}{{\text{y}}}$ and Q = 2y. As, we know that $\int {\dfrac{1}{{\text{x}}}{\text{dx}}} {\text{ = lnx}}$ and ${{\text{e}}^{\ln {\text{x}}}}{\text{ = x}}$. Also, $\log {{\text{m}}^{\text{n}}}{\text{ = n(logm)}}$.
So, ${\text{I}}{\text{.F = }}{{\text{e}}^{\int {{\text{Pdy}}} }}{\text{ = }}{{\text{e}}^{\int { - \dfrac{{{\text{dy}}}}{{\text{y}}}} }}{\text{ = }}{{\text{e}}^{ - \ln {\text{y}}}}{\text{ = }}{{\text{e}}^{\ln \dfrac{1}{{\text{y}}}}}{\text{ }}$
Therefore, I.F = $\dfrac{1}{{\text{y}}}$
Therefore, solution is ${\text{x(}}\dfrac{1}{{\text{y}}}){\text{ = }}\int {{\text{2y(}}\dfrac{1}{{\text{y}}}){\text{dy}}}$
${\text{x(}}\dfrac{1}{{\text{y}}}){\text{ = }}\int {{\text{2dy}}}$
${\text{x(}}\dfrac{1}{{\text{y}}}){\text{ = 2y + C}}$
So, we get ${\text{x = 2}}{{\text{y}}^2} + {\text{ Cy}}$.
Now, we have f (-1) = 1
$1{\text{ = 2( - 1}}{{\text{)}}^2}{\text{ + C( - 1)}}$
1 = 2 – C
C = 1
Applying value of C, we get
${\text{x = 2}}{{\text{y}}^2}{\text{ + y}}$
Now, we have to find the value at f (1). Therefore, putting y = 1 in the above equation, we get
${\text{x = 2(1}}{{\text{)}}^2}{\text{ + 1}}$
x = 3
so, option (B) is correct.

Note: When we come up with such types of questions, we will require some knowledge of integration to find the solution of the given differential equation. First, we will find the solution of the differential equation and then we will apply the values provided in the question to find the integration constant. After it, we apply that value in the function, whose value we have to find.