Question

Solution of the equation x²y – x³$\frac{dy}{dx}$ = y⁴cos x, when y (0) = 1, is –

• A. y³ = 3x³ sin x
• B. x³ = 3y³ sin x
• C. x³ = y³ sin x
• D. None of these

Answer 1

Answer: (B)

x³ = 3y³ sin x

Answer 2

We have, x²y – x³ $\frac{dy}{dx}$ = y⁴ cos x

Ž x³$\frac{dy}{dx}$ – x²y = –y⁴ cos x

Dividing by –y⁴ x³, we get

– $\frac{1}{y^{4}}$ $\frac{dy}{dx}$ + $\frac{1}{y^{3}}$ . $\frac{1}{x}$ = $\frac{1}{x^{3}}$ cos x

On putting $\frac{1}{y^{3}}$ = V so that – $\frac { 1 } { y ^ { 4 } }$ $\frac{dy}{dx}$ = $\frac { 1 } { 3 }$ $\frac{dv}{dx}$, we get

$\frac{1}{3}$ $\frac{dV}{dx}$ + $\frac{1}{x}$V = $\frac{1}{x^{3}}$ cos x

Ž $\frac{dV}{dx}$ + $\frac{3}{x}$V = $\frac{3}{x^{3}}$cos x.

which is linear in V.

I. F. = $e^{\int_{}^{}{\frac{3}{x}dx}}$ = e^{3 log x} = x³.

So the solution is

x³V = $\int_{}^{}x^{3}$ . $\frac{3}{x^{3}}$ cos x dx + c = 3 sin x + c.

Ž $\frac{x^{3}}{y^{3}}$ = 3 sin x + c.

On putting x = 0, y = 1, we get c = 0.

Hence the solution is x³ = 3y³ sin x