Question

‌‌How‌ ‌do‌ ‌I‌ ‌solve‌ ‌the‌ ‌differential‌ ‌equation‌ ‌$y'+{{x}^{2}}y={{x}^{2}}$?‌ ‌

Answer 1

Hint: Now we are given with a differential equation. To solve the equation we will separate the variables x and y. Then we will Integrate the equation on both sides. Now using the general integral we will solve the integral and hence find the solution of the given differential equation.

Complete step by step solution:
Consider the given equation $y'+{{x}^{2}}y={{x}^{2}}$
Now we know that $y'=\dfrac{dy}{dx}$ Hence the equation is,
$\Rightarrow \dfrac{dy}{dx}+{{x}^{2}}y={{x}^{2}}$
We will solve the above differential equation by Variable separable method, Hence we will separate the variables x and y and solve the equation
Now we will try to separate the x and y variables. To do so first let us transpose ${{x}^{2}}y$ to RHS. Hence we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}={{x}^{2}}-{{x}^{2}}y \\
 & \Rightarrow \dfrac{dy}{dx}={{x}^{2}}\left( 1-y \right) \\
\end{align}$
Now dividing the equation by $\left( 1-y \right)$ and cross multiplying the terms we get,
$\Rightarrow \dfrac{dy}{\left( 1-y \right)}={{x}^{2}}dx$
Now let us integrate the equation on both sides. Hence we get,
$\Rightarrow \int{\dfrac{dy}{1-y}}=\int{{{x}^{2}}dx}$
Now let us solve the Integral on LHS.
Consider $\int{\dfrac{dy}{1-y}}$
We will solve the integral by method of substitution. Let $1-y=t$ . Now differentiating the equation we get $-dy=dt$ .
Now let us substitute the values in the above equation.
$\begin{align}
  & \Rightarrow \int{\dfrac{dy}{1-y}}=\int{\dfrac{-dt}{t}} \\
 & \Rightarrow \int{\dfrac{dy}{1-y}}=-\int{\dfrac{dt}{t}} \\
 & \Rightarrow \int{\dfrac{dy}{1-y}}=-\ln t+C \\
 & \Rightarrow \int{\dfrac{dy}{1-y}}=-\ln \left( 1-y \right)+C \\
\end{align}$
Now consider the integral $\int{{{x}^{2}}dx}$ .
We know that integration of ${{x}^{n}}$ is given by $\dfrac{{{x}^{n+1}}}{n+1}$ .
Hence we have $\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C$
Now substituting the values of the integral in the equation $\int{\dfrac{dy}{1-y}}=\int{{{x}^{2}}dx}$ we get,
$\Rightarrow -\ln \left( 1-y \right)=\dfrac{{{x}^{3}}}{3}+C'$ where C’ is a constant.
Hence the solution of the given differential equation is $-\ln \left( 1-y \right)=\dfrac{{{x}^{3}}}{3}+C'$ .

Note: Now note that when we are integrating an expression without limits then always add the constant of Integral. Also while using Method of substitution remember to substitute the differential coefficient dx as well.