Question

How do you solve $y'=-xy+\sqrt{y}$ given $y\left( 0 \right)=1$?

Answer 1

Hint: We need to divide the given equation by $\sqrt{y}$ to get \[\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}=-x\sqrt{y}+1\]. Then on substituting $t=\sqrt{y}$ we will obtain the equation \[2\dfrac{dt}{dx}+xt=1\], which can be solved by finding the integrating factor. Finally, on substituting the solution in $y={{t}^{2}}$, we will obtain the solution in terms of a constant. The value of the constant can be found by using $y\left( 0 \right)=1$, which has been given in the question.

Complete step by step solution:
The differential equation given in the above question is written as
$\Rightarrow y'=-xy+\sqrt{y}$
Writing $y=\dfrac{dy}{dx}$ in the above equation, we get
$\Rightarrow \dfrac{dy}{dx}=-xy+\sqrt{y}$
Dividing by $\sqrt{y}$ both the sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}=\dfrac{1}{\sqrt{y}}\left( -xy+\sqrt{y} \right) \\
 & \Rightarrow \dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}=-x\sqrt{y}+1.........\left( i \right) \\
\end{align}\]
Now, let us substitute $t=\sqrt{y}$ so that
$\Rightarrow t=\sqrt{y}........\left( ii \right)$
Differentiating both the sides with respect to x, we get
$\Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}$
Multiplying \[2\] on both the sides, we get
\[\begin{align}
  & \Rightarrow 2\dfrac{dt}{dx}=\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}=2\dfrac{dt}{dx}......\left( iii \right) \\
\end{align}\]
On substituting the equations (ii) and (iii) into the equation (i) we get
\[\Rightarrow 2\dfrac{dt}{dx}=-xt+1\]
Adding $xt$ both the sides, we get
\[\Rightarrow 2\dfrac{dt}{dx}+xt=1\]
Dividing the above equation by $2$ we get
\[\Rightarrow \dfrac{dt}{dx}+\dfrac{x}{2}t=\dfrac{1}{2}.......\left( iv \right)\]
The integrating factor for the above differential equation can be given by
$\begin{align}
  & \Rightarrow IF={{e}^{\int{\dfrac{x}{2}dx}}} \\
 & \Rightarrow IF={{e}^{\dfrac{{{x}^{2}}}{4}}} \\
\end{align}$
Multiplying both the sides of (iv) by the integrating factor ${{e}^{\dfrac{{{x}^{2}}}{4}}}$, we get
\[\begin{align}
  & \Rightarrow 2{{e}^{\dfrac{{{x}^{2}}}{4}}}\dfrac{dt}{dx}+{{e}^{\dfrac{{{x}^{2}}}{4}}}\left( xt \right)={{e}^{\dfrac{{{x}^{2}}}{4}}} \\
 & \Rightarrow \dfrac{d\left( t{{e}^{\dfrac{{{x}^{2}}}{4}}} \right)}{dx}={{e}^{\dfrac{{{x}^{2}}}{4}}} \\
 & \Rightarrow t{{e}^{\dfrac{{{x}^{2}}}{4}}}=\int{{{e}^{\dfrac{{{x}^{2}}}{4}}}dx}.........\left( v \right) \\
\end{align}\]
Now, we know that
\[\Rightarrow {{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{6}+\dfrac{{{x}^{4}}}{24}+.......\]
Replacing $x$ by $\dfrac{{{x}^{2}}}{4}$ we get
\[\begin{align}
  & \Rightarrow {{e}^{\dfrac{{{x}^{2}}}{4}}}=1+\dfrac{{{x}^{2}}}{4}+\dfrac{{{\left( \dfrac{{{x}^{2}}}{4} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{{{x}^{2}}}{4} \right)}^{3}}}{6}+....... \\
 & \Rightarrow {{e}^{\dfrac{{{x}^{2}}}{4}}}=1+\dfrac{{{x}^{2}}}{4}+\dfrac{{{x}^{4}}}{32}+\dfrac{{{x}^{6}}}{384}+....... \\
\end{align}\]
Substituting the above in the equation (v) we get
\[\begin{align}
  & \Rightarrow t\left( {{e}^{\dfrac{{{x}^{2}}}{4}}} \right)=\int{\left( 1+\dfrac{{{x}^{2}}}{4}+\dfrac{{{x}^{4}}}{32}+\dfrac{{{x}^{6}}}{384}+....... \right)dx} \\
 & \Rightarrow t\left( {{e}^{\dfrac{{{x}^{2}}}{4}}} \right)=x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+.......+C \\
\end{align}\]
Dividing both the sides by ${{e}^{\dfrac{{{x}^{2}}}{4}}}$ we get
\[\Rightarrow t={{e}^{-\dfrac{{{x}^{2}}}{4}}}\left( x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+....... \right)+C{{e}^{-\dfrac{{{x}^{2}}}{4}}}........\left( vi \right)\]
Now, from the equation (ii) we have
$\Rightarrow t=\sqrt{y}$
Considering $x=0$ both the sides, we get
$\Rightarrow t\left( 0 \right)=\sqrt{y\left( 0 \right)}$
According to the question, we have $y\left( 0 \right)=1$. So we get
$\begin{align}
  & \Rightarrow t\left( 0 \right)=\sqrt{1} \\
 & \Rightarrow t\left( 0 \right)=1.......\left( vii \right) \\
\end{align}$
Substituting $x=0$ in the above equation (vi) we get
\[\begin{align}
  & \Rightarrow t\left( 0 \right)={{e}^{-\dfrac{{{\left( 0 \right)}^{2}}}{4}}}\left( \left( 0 \right)+\dfrac{{{\left( 0 \right)}^{3}}}{12}+\dfrac{{{\left( 0 \right)}^{5}}}{160}+\dfrac{{{\left( 0 \right)}^{7}}}{2688}+....... \right)+C{{e}^{-\dfrac{{{\left( 0 \right)}^{2}}}{4}}} \\
 & \Rightarrow t\left( 0 \right)=C \\
\end{align}\]
Substituting (vii) we get
$\begin{align}
  & \Rightarrow 1=C \\
 & \Rightarrow C=1 \\
\end{align}$
Substituting this in the equation (vi) we get
\[\begin{align}
  & \Rightarrow t={{e}^{-\dfrac{{{x}^{2}}}{4}}}\left( x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+....... \right)+{{e}^{-\dfrac{{{x}^{2}}}{4}}} \\
 & \Rightarrow t={{e}^{-\dfrac{{{x}^{2}}}{4}}}\left( 1+x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+....... \right).........\left( vii \right) \\
\end{align}\]
On squaring both the sides of equation (ii), we get
$\begin{align}
  & \Rightarrow {{t}^{2}}={{\sqrt{y}}^{2}} \\
 & \Rightarrow {{t}^{2}}=y \\
 & \Rightarrow y={{t}^{2}} \\
\end{align}$
Substituting equation (vii) in the above equation, we get
\[\begin{align}
  & \Rightarrow y={{\left[ {{e}^{-\dfrac{{{x}^{2}}}{4}}}\left( 1+x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+....... \right) \right]}^{2}} \\
 & \Rightarrow y={{e}^{-\dfrac{{{x}^{2}}}{4}\times 2}}{{\left( 1+x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+....... \right)}^{2}} \\
 & \Rightarrow y={{e}^{-\dfrac{{{x}^{2}}}{2}}}{{\left( 1+x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+....... \right)}^{2}} \\
\end{align}\]
Hence, we have finally obtained the solution of the given differential equation as \[y={{e}^{-\dfrac{{{x}^{2}}}{2}}}{{\left( 1+x+\dfrac{{{x}^{3}}}{12}+\dfrac{{{x}^{5}}}{160}+\dfrac{{{x}^{7}}}{2688}+....... \right)}^{2}}\]

Note: We must remember the Taylor series expansion of the functions in order to solve these types of questions. This is because the integration \[\int{{{e}^{\dfrac{{{x}^{2}}}{4}}}dx}\] was unsolvable, and hence we obtained the solution in the form of the series. Do not forget to obtain y as a function of x from the obtained solution for t, since the original variable was y and not t.