Question

solve x^2d^2y/dx^2+4xdy/dx+2y=e^x​

Answer 1

SOLUTION

TO SOLVE

$\displaystyle \sf{ {x}^{2} \frac{ {d}^{2}y }{d {x}^{2} } + 4x \frac{dy}{dx} + 2y = {e}^{x} }$

EVALUATION

Here the given differential equation is

$\displaystyle \sf{ {x}^{2} \frac{ {d}^{2}y }{d {x}^{2} } + 4x \frac{dy}{dx} + 2y = {e}^{x} }$

Let us assume that

$\displaystyle \sf{ x = {e}^{t} }$

$\displaystyle \sf{ \implies \: t = \log x }$

So the given differential equation becomes

$\displaystyle \sf{ \bigg[ \: D(D - 1) + 4D + 2 \bigg]y = {e}^{ {e}^{t} } }$

On simplification we get

$\displaystyle \sf{ \bigg[ \: {D}^{2} + 3D + 2 \bigg]y = {e}^{ {e}^{t} } }$

Let

$\displaystyle \sf{ y = {e}^{ mt } }$

be the trial solution

Then the auxiliary equation is

$\sf{ {m}^{2} + 3m + 2 = 0 }$

$\displaystyle \sf{ \implies \: m = - 1 \: ,\: - 2 }$

So the complementary function

= C. F

$\displaystyle \sf{ = a \: {e}^{ - t} + b \: {e}^{ - 2t} }$

$\displaystyle \sf{ = a \: {e}^{ - \log x} + b \: {e}^{ - 2 \log x} }$

$\displaystyle \sf{ = a \: {x}^{ - 1} + b \: {x}^{ - 2} }$

Where a and b are constants

Now the particular integral

= P. I

$\displaystyle \sf{ = \: \frac{1}{( \: {D}^{2} + 3D + 2)} \: {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = \: \frac{1}{( \: D + 1)(D + 2)} \: {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = \frac{1}{( \: D + 1)} \: {e}^{ {e}^{t} } - \frac{1}{(D + 2)} \: {e}^{ {e}^{t} } }$

Now

$\displaystyle \sf{ \frac{1}{( \: D + 1)} \: {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = \frac{1}{( \: D + 1)} \: {e}^{ - t} . {e}^{t} . {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = {e}^{ - t} . \frac{1}{( \: D - 1 + 1)} \: {e}^{t} . {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = {e}^{ - t} . \frac{1}{ D } \: {e}^{t} . {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = {e}^{ - t} . \int {e}^{ {e}^{t} } d({e}^{t}) }$

$\displaystyle \sf{ = {x}^{ - 1} . \int {e}^{x} dx }$

$\displaystyle \sf{ = {x}^{ - 1} \: {e}^{x} }$

Again

$\displaystyle \sf{ \frac{1}{( \: D + 2)} \: {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = \frac{1}{( \: D + 2)} \: {e}^{ - 2t} . {e}^{2t} . {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = {e}^{ - 2t} . \frac{1}{( \: D - 2 + 2)} \: {e}^{2t} . {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = {e}^{ - 2t} . \frac{1}{ D } \: {e}^{2t} . {e}^{ {e}^{t} } }$

$\displaystyle \sf{ = {e}^{ - t} . \int {e}^{t}. {e}^{ {e}^{t} } d({e}^{t}) }$

$\displaystyle \sf{ = {x}^{ - 2} . \int x.{e}^{x} dx }$

$\displaystyle \sf{ = {x}^{ - 2} ( x.{e}^{x} - {e}^{x} )}$

Thus particular integral

$\displaystyle \sf{ = {x}^{ - 1} \: {e}^{x} - {x}^{ - 2} ( x.{e}^{x} - {e}^{x} ) }$

$\displaystyle \sf{ = {x}^{ - 1} \: {e}^{x} - {x}^{ - 1} \: {e}^{x} + {x}^{ - 2} {e}^{x} }$

$\displaystyle \sf{ = {x}^{ - 2} {e}^{x} }$

Hence the required required solution

= C. F + P. I

$\displaystyle \sf{ = a \: {x}^{ - 1} + b \: {x}^{ - 2} + {x}^{ - 2} \: {e}^{x} }$

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Answer 2

Step-by-step explanation:

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