Home
Class 12
MATHS
Solution of y dx – x dy = x^2 ydx is:...

Solution of y dx – x dy = `x^2 `ydx is:

A

`ye ^(x^2)= cx^2`

B

`ye ^(-x^2)= cx^2`

C

`Y^2e^(x^2) = cx^2`

D

`y^2 e^(-x^2)= cx^2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the differential equation given by \( y \, dx - x \, dy = x^2 y \, dx \), we will follow these steps: ### Step 1: Rearranging the Equation We start with the equation: \[ y \, dx - x \, dy = x^2 y \, dx \] We can rearrange this by moving all terms involving \( dy \) to one side and terms involving \( dx \) to the other side: \[ y \, dx - x^2 y \, dx = x \, dy \] Factoring out \( dx \) from the left side gives: \[ (y - x^2 y) \, dx = x \, dy \] ### Step 2: Dividing by \( dx \) Next, we divide both sides by \( dx \): \[ y - x^2 y = x \frac{dy}{dx} \] This simplifies to: \[ y(1 - x^2) = x \frac{dy}{dx} \] ### Step 3: Separating Variables Now we can separate variables: \[ \frac{dy}{y} = \frac{1 - x^2}{x} \, dx \] ### Step 4: Integrating Both Sides We will now integrate both sides: \[ \int \frac{dy}{y} = \int \left( \frac{1}{x} - x \right) \, dx \] The left side integrates to: \[ \log |y| \] The right side can be integrated as follows: \[ \int \frac{1}{x} \, dx - \int x \, dx = \log |x| - \frac{x^2}{2} + C \] ### Step 5: Combining Results Putting it all together, we have: \[ \log |y| = \log |x| - \frac{x^2}{2} + C \] We can express this in a more convenient form: \[ \log |y| - \log |x| = -\frac{x^2}{2} + C \] This can be rewritten using properties of logarithms: \[ \log \left| \frac{y}{x} \right| = -\frac{x^2}{2} + C \] ### Step 6: Exponentiating Both Sides To eliminate the logarithm, we exponentiate both sides: \[ \frac{y}{x} = e^{-\frac{x^2}{2} + C} \] Let \( k = e^C \), then: \[ y = kx e^{-\frac{x^2}{2}} \] ### Final Solution Thus, the solution to the differential equation is: \[ y = kx e^{-\frac{x^2}{2}} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

Solution of x (dy)/(dx) + y = x^(2) y^(4) is

The solution of the diffferential equation x (dy)/(dx ) + 2y = x^(2) is

The general solution of (dy)/(dx) = 2x e^(x^(2)-y) is

General solution of (y+y^(2))dy=(x+x^(2))dx is

The solution of ydx-xdy+(1+x^(2))dx+x^(2)siny dy=0, is given by

Let the solution of y log ydx+(x-log y)dy=0 is x.log y=A(log y)^(2)+c where c is constant then A=?