Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

1. $y=e^x+1$ : $y^{\prime \prime }-y=0$
2. $y=x^2+2x+C$ : $y^{\prime }-2x-2=0$
3. $y=\cos x+C$ : $y^{\prime }+\sin x=0$
4. $y=\sqrt{1+x^2}$ : $y^{\prime }=\frac{xy}{1+x^2}$
5. $y=Ax$ : $xy=y(x\ne 0)$
6. $y=x\sin x$ : $xy=y+x\sqrt{x^2{y}^2}(x\ne 0andx>yorx<y)$
7. $xy=\log y+C$ : $y^{\prime }=\frac{y^2}{1-xy}(xy\ne 1)$
8. $y-\cos y=x$ : $(y\sin y+\cos y+x)y^{\prime }=y$
9. $x+y={\tan }^{-1}y$ : $y^2{y}^{\prime }+y^2+1=0$
10. $y=\sqrt{a^2-x^2}xϵ(-a,a)$ : $x+y\frac{dy}{dx}=0(y\ne 0)$

Answer 1

1) Given, $y=e^x+1$; $y^{\prime \prime }-y=0$

For ex. $y^{\prime \prime }-y=0\Rightarrow \frac{d^{2}y}{d{x}^2}–y=0$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=y$ …(1)

For this type of equation solunation is

$y=e^x+c$ $C\in R$

And it satisfies the above eqn.

So, $y=e^x+1$

2) $y^{\prime }-2x-2=0$

$\Rightarrow \frac{dy}{dx}=2x+2$

$\Rightarrow dy=(2x+2)dx$

Integrating on both sides, we get

$\int dy=\int (2x+2)dx$

$y=x^2+2x+c,C\in R$

3) $y^{\prime }+\sin x=0$

$\Rightarrow \frac{dy}{dx}+\sin =0$

$\Rightarrow dy=\sin dx$

Integrating on both side

$\int dy=\int \sin xdx$

$\Rightarrow y=+\cos x+c$ $\{\int \sin xdx=\cot x\}$

4) $y^{\prime }=\frac{xy}{1+x^2}$

$\Rightarrow \frac{dy}{dx}=\frac{xy}{1=x^2}$

$\Rightarrow \frac{dy}{y}=\frac{xdx}{1+x^2}$

$\Rightarrow \int \frac{dy}{y}=\int \frac{xdx}{1+x^2}$

Let $x^2+1=t$

$2xdx=dt\pm xdx=\frac{dt}{2}$

$\Rightarrow \ln \left(y\right)=\frac{1}{2}$ $\int \frac{dt}{\left(1pt\right)}$

$\Rightarrow \ln \left(y\right)=\frac{1}{2}\left[\ln \right(1+2\left)\right]+c,c\in R$

$\Rightarrow$ taking $c=0$

$\ln \left(y\right)=\frac{1}{2}\ln (1+1)=\ln (1+1{)}^{1/2}$

$y=(1+1{)}^{1/2}$

$y=\sqrt{1+x^2}$

5) $y=Ax$; $xy=y(x\ne 0)$

$x\frac{dy}{dx}=y\Rightarrow \frac{dy}{y}=\frac{dx}{x}\Rightarrow \int \frac{dy}{y}=\int \frac{dx}{x}$

$\ln \left(y\right)=\ln \left(x\right)+\ln \left(A\right)\Rightarrow \ln \left(y\right)=\ln \left(Ax\right),A\in R$

$\Rightarrow y=Ax$

9) $x+y={\tan }^{-1}y$; $y^2{y}^{\prime }+y^2+1=0$

$y^2{y}^1=-(1+y^2)$

$y^1=\frac{-(1+y^2)}{y^2}$

$\frac{y^2{y}^1}{1+y^2}=(-1)$

$\int \frac{y^2}{1+y^2}dy=\int –dx$

$\Rightarrow \int \frac{1+y^2-1}{1+y^2}dy=-\int dx$

$=\int dy-\int \frac{1}{1+y^2}dy=-\int dx$

$\Rightarrow y-{\tan }^{-1}\left(y\right)=-x+c$

$\Rightarrow x+y={\tan }^{-1}\left(y\right)+c$

10) $y=\sqrt{a^2-x^2}xϵ(-a,a)$; $x+y\frac{dy}{dx}=0(y\ne 0)$

$x+y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-x}{y}$

$\int ydy=\int –xdx$

$\frac{y^2}{2}=\int –xdx$

$y^2=a-x^2$

$y=\sqrt{a^2-x^2},xϵ(-a,a)$