Question

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upcoming exam

Answer 1

The first equation is linear:


`x(\mathrm dy)/(\mathrm dx)-y=x^2\sin x`

Divide through by
`x^2` to get


`\frac1x(\mathrm dy)/(\mathrm dx)-\frac1{x^2}y=\sin x`

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for
`y`.


`(\mathrm d)/(\mathrm dx)\left[\frac1xy\right]=\sin x`

`\implies\frac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C`

`\implies y=-x\cos x+Cx`

- - -

The second equation is also linear:


`x^2y'+x(x+2)y=e^x`

Multiply both sides by
`e^x` to get


`x^2e^xy'+x(x+2)e^xy=e^(2x)`

and recall that
`(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x`, so we can write


`(x^2e^xy)'=e^(2x)`

`\implies x^2e^xy=\displaystyle\int e^(2x)\,\mathrm dx=\frac12e^(2x)+C`

`\implies y=(e^x)/(2x^2)+\frac C{x^2e^x}`

- - -

Yet another linear ODE:


`\cos x(\mathrm dy)/(\mathrm dx)+\sin x\,y=1`

Divide through by
`\cos^2x`, giving


`\frac1{\cos x}(\mathrm dy)/(\mathrm dx)+(\sin x)/(\cos^2x)y=\frac1{\cos^2x}`

`\sec x(\mathrm dy)/(\mathrm dx)+\sec x\tan x\,y=\sec^2x`

`(\mathrm d)/(\mathrm dx)[\sec x\,y]=\sec^2x`

`\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C`

`\implies y=\cos x\tan x+C\cos x`

`y=\sin x+C\cos x`

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation


`a(x)y'(x)+b(x)y(x)=c(x)`

then rewrite it as


`y'(x)=(b(x))/(a(x))y(x)=(c(x))/(a(x))\iff y'(x)+P(x)y(x)=Q(x)`

The integrating factor is a function
`\mu(x)` such that


`\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'`

which requires that


`\mu(x)P(x)=\mu'(x)`

This is a separable ODE, so solving for
`\mu` we have


`\mu(x)P(x)=(\mathrm d\mu(x))/(\mathrm dx)\iff(\mathrm d\mu(x))/(\mu(x))=P(x)\,\mathrm dx`

`\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx`

`\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)`

and so on.