# Find y as a function of x if (x^2)*y double prime + 2x*y prime - 30y = x^6, y(1) = 6, y prime (1)...

## Question:

Find {eq}y {/eq} as a function of {eq}x {/eq} if {eq}\; x^2 {y}'' + 2x{y}' - 30y = x^6, \; y(1) = 6, \; {y}'(1) = -5 {/eq}.

## The Initial Value Problem in Calculus:

The second order differential equation of the type {eq}z^2 y''+p_{1}xy'+p_{2}y= 0 {/eq} , where {eq}p_{1}, p_{2} {/eq} are arbitrary constants.

It may be transformed into a linear equation with constant coefficients by means of the substitution {eq}\dfrac {dt}{dx}=\dfrac {1}{x} {/eq} , where {eq}x=e^t {/eq} .

Now {eq}\dfrac {dy}{dx}=\dfrac {dy}{dt}\cdot \dfrac {dt}{dx}=\dfrac {1}{x}\cdot \dfrac {dy}{dt} {/eq}, that is, {eq}x\cdot \dfrac {dy}{dx}=\dfrac {dy}{dt}{/eq} or, {eq}xDt=\theta y {/eq} or, {eq}ty'=\theta y {/eq}

Similarly, {eq}x^{2}\dfrac {d^{2}y}{dx^{2}}=\dfrac {d^{2}y}{dt^{2}}-\dfrac {dy}{dt}=\left( \theta ^{2}-\theta \right) y {/eq}, that is {eq}x^{2}D^{2}y=\left( \theta -1\right) \theta y {/eq} or, {eq}t^{2}y''=\left( \theta -1\right) \theta y {/eq}

Now this problem is an initial value problem. The solution of the problem always satisfies the initial conditions.

## Answer and Explanation:

We are given: {eq}x^2 {y}'' + 2x{y}' - 30y = x^6 {/eq}

Proceeding as above {eq}\left(x=e^{t},t=\log x,t\dfrac {d}{dx}=\dfrac {d}{dt}= \theta...

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