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Gre4nikov [31]
3 years ago
15

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

oming exam

Mathematics
1 answer:
mariarad [96]3 years ago
6 0
The first equation is linear:

x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x

Divide through by x^2 to get

\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{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.

\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x
\implies\dfrac1xy=\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=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}

- - -

Yet another linear ODE:

\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1

Divide through by \cos^2x, giving

\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}
\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x
\dfrac{\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)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{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)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\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.
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